Expert: Paul Klarreich - 4/6/2008

Question
A plane has equation x - 2y +z = 4.  Show that the square, say D^2, of the distance from the point P=(-1,3,2) to a point (x,y,z)on the plane is given by
D^2 = 2x^2 - 2x +14 + 5y^2 + 2y - 4xy.j
By finding the minimum of D^2 determine the shortest distance form P to the plane.  (You should verify that the distance is indeed a minimum).  Let Q be the point on the plane clasest to P.  Show that the line QP is perpendicular to the plane at Q.

Answer
Questioner:   Natalie
Category:  Calculus
Private:  No
 
Subject:  minimum in multivariable calculus
Question:  A plane has equation x - 2y +z = 4.  Show that the square, say D^2, of the distance from the point P=(-1,3,2) to a point (x,y,z)on the plane is given by
D^2 = 2x^2 - 2x +14 + 5y^2 + 2y - 4xy.j
By finding the minimum of D^2 determine the shortest distance form P to the plane.  (You should verify that the distance is indeed a minimum).  Let Q be the point on the plane closest to P.  Show that the line QP is perpendicular to the plane at Q.
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Hi, Natalie,

If P=(-1,3,2) then your D^2 to a point (x,y,z) would be

(x + 1)^2 + (y - 3)^2 + (z - 2)^2  

by the distance formula.  Expand that:

x^2 + 2x + 1 + y^2 - 6y + 9 + z^2 - 4z + 4

x^2 + 2x  + y^2 - 6y + z^2 - 2z + 14

Now that's the D^2 from P to any old point.  But if the point is on the plane, then you can use the equation of your plane to eliminate a variable.  Eliminate z by writing:

x - 2y +z = 4

z = 4 + 2y - x

and substituting:

x^2 + 2x  + y^2 - 6y + z^2 - 2z + 14
x^2 + 2x  + y^2 - 6y + ()^2 - 2() + 14
x^2 + 2x  + y^2 - 6y + (4 + 2y - x)^2 - 4(4 + 2y - x)+ 14

Expand again.

x^2 + 2x  + y^2 - 6y + 16 + 4y^2 + x^2 + 16y - 8x - 4xy - 16 - 8y + 4x + 14

Combine stuff:

2x^2 - 2x + 5y^2 + 2y  - 4xy    + 14

OK, that is your D^2, which we call big-D. [I hope you are not from Dallas.]

Now for your minimum.  You will find Dx and Dy:

Dx = 4x - 2 - 4y
Dy = 10y + 2 - 4x.

Now set those both equal to zero and solve.

4x - 2 - 4y = 0
10y + 2 - 4x = 0
---------------------- << add
6y = 0

y = 0

4x - 2 = 0
x = 1/2

Now use z = 4 + 2y - x to find z.

z = 4 + 2(0) - 1/2 = 7/2

Your point Q is (1/2, 0, 7/2)

[OK, you have to do some second derivative stuff to verify that it is a minimum, but I'll leave that to you.]

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Now to show that the line PQ is perp to the plane use this fact:

If the equation of the plane is  ax + by + cz = d, then <a,b,c> is  a vector normal to the plane.

Now P(-1,3,2)  Q(1/2, 0,7/2) form a vector PQ<3/2,-3,3/2>

And the plane  x - 2y +z = 4  has a normal vector N<1,-2,1>.  You can easily confirm that  (3/2)N = PQ.  So PQ is in the direction of N and is perp to the plane.